1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauss theorem and the nature of the surface

  1. Apr 24, 2013 #1
    According to Gauss’ law the total number of lines of force over a closed surface is equal to 1/ε times the net charge enclosed within the closed surface. Why should it be a closed surface but not an open surface too? I am unable to find a convincing explanation for it. Since we take into account only the number of lines starting or reaching a charge, the total number of lines is not going to vary whether we take a closed surface or open surface near the charge. Then why should it be specifically a closed one? Can anyone giving a convincing explanation?
     
  2. jcsd
  3. Apr 24, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    If the surface is open, you can get any arbitrary value. It is important that you enclose the whole mass - without leaks, so to speak.
    There is no "number of lines of force".
     
  4. Apr 25, 2013 #3
    Thank you for the reply. If there is no lines of force, then how the calculated value of 1.129 x 1011 lines of force from 1 C of charge placed in air or vacuum arrived? What does that number exactly stand for? The value has been obtained from the formula 1/ε for air.
     
  5. Apr 25, 2013 #4

    sz0

    User Avatar

    I am by means an expert but, what Gaus law does is to collect all that flows through a surface. If you dont totally enclose the charge there will be flow going out that you are not calculating.
     
  6. Apr 25, 2013 #5
    Because only a closed surface can enclose a charge.

    How do you define "enclosed charge" when your surface is open?
     
  7. Apr 25, 2013 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I have no idea who did that, but it is wrong.
    Nothing.
    $$\frac{1C}{\epsilon_0}=1.13\times 10^{11} Vm$$
    The numerical value depends on the units you use - if you convert this to imperial units, you get a different number, for example. This alone shows that the numerical value itself cannot have a physical meaning (like some number of "lines").
     
  8. Apr 26, 2013 #7
     
  9. Apr 26, 2013 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It is pointless to talk about a "number of lines of force". Such a thing does not exist.
    I think "line of force" itself is a problematic concept, but at least it has some clear meaning (=field lines). Let's look at the points 1m away from a charge: Every point is on its own "line of force", and the number of points in a distance of 1m is infinite. There is no way to get any finite number for "lines of force". If you want to draw them, you have to restrict yourself to a finite number, but that is not an exact drawing of the field.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gauss theorem and the nature of the surface
  1. The Nature of IS (Replies: 0)

  2. Gauss's Law (Replies: 2)

Loading...